S5 Biology Practice Paper: Plant Nutrition, Photosynthesis And Water Transport (12 Questions, 100 Marks, 75 Minutes)

S5 Biology Practice Paper: Plant Nutrition, Photosynthesis and Water Transport (12 Questions, 100 Marks, 75 Minutes)

2026-11-09

S5 Biology: Photosynthesis, Transpiration and Plant Transport

S5 biology: photosynthesis, leaf adaptations, stomata, transpiration pull, xylem and phloem transport, mineral deficiency and limiting factors. 12 questions with answers.

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Instructions: This paper has three sections, A, B and C. Answer all questions. Total marks: 100. Time allowed: 75 minutes.

Show the formula used, the values substituted and the correct units. Give numerical answers to three significant figures.

In explanations, name the principle used, for example limiting factors, cohesion and adhesion, or source and sink.

Spend about 20 minutes on Section A, 30 minutes on Section B and the last 25 minutes on Section C.

Section A: Photosynthesis and Leaf Structure (Questions 1-4)

Four questions, 28 marks in total. Suggested time: 20 minutes.

#	Question	Answer and marking notes
1	Write the overall equation for photosynthesis using chemical symbols, and state where the light-dependent stage and the Calvin cycle take place in a chloroplast. (6 marks)	6CO2 + 6H2O -> C6H12O6 + 6O2 (light energy and chlorophyll required): 2 marks for correct reactants and products. The light-dependent stage occurs on the thylakoid (grana) membranes: 2 marks. The Calvin cycle occurs in the stroma: 2 marks.
2	Describe three structural adaptations of a leaf for photosynthesis, giving one reason for each. (6 marks)	A thin transparent cuticle on the upper epidermis lets light pass to the mesophyll: 2 marks. Palisade cells are closely packed with many chloroplasts, giving a large light-absorbing surface: 2 marks. Large air spaces in the spongy mesophyll speed up diffusion of carbon dioxide and oxygen: 1 mark. Stomata are mainly on the lower surface, reducing water loss while still allowing gas exchange: 1 mark.
3	A plant absorbs 2.24 dm3 of carbon dioxide in 24 hours (molar volume of gas at room temperature = 24 dm3/mol). (a) Calculate the number of moles of carbon dioxide absorbed. (b) Assuming all of it is used in photosynthesis, calculate the mass of glucose produced (molar mass 180 g/mol). (c) State the volume of oxygen released. (8 marks)	(a) n(CO2) = 2.24 / 24 = 0.0933 mol: 3 marks (formula, substitution, answer). (b) n(glucose) = 0.0933 / 6 = 0.0156 mol: 2 marks; mass = 0.0156 x 180 = 2.80 g: 2 marks. (c) The equation shows 6 mol of carbon dioxide give 6 mol of oxygen, so the volume of oxygen is also 2.24 dm3: 1 mark.
4	A leaf loses 12.0 g of water in 6.0 hours. Its surface area is 0.0150 m2. (a) Calculate the transpiration rate in g m-2 h-1. (b) Predict the effect of coating the lower surface with petroleum jelly and explain your answer. (8 marks)	(a) Rate = 12.0 / (6.0 x 0.0150) = 133 g m-2 h-1: 4 marks (formula, substitution, answer with unit to three significant figures). (b) The rate falls clearly: 2 marks, because most stomata are on the lower surface; the petroleum jelly blocks them so water vapour cannot diffuse out: 2 marks.

Section B: Transpiration, Transport and Mineral Nutrition (Questions 5-8)

Four questions, 32 marks in total. Suggested time: 30 minutes.

#	Question	Answer and marking notes
5	Describe how the transpiration pull moves water from the root to the leaves, and state two factors that keep the water column unbroken. (8 marks)	Transpiration from the leaf lowers the water potential of the mesophyll cells, so water is drawn from the xylem, creating a negative pressure (tension) in the xylem: 2 marks; this tension is transmitted down the water column and pulls water up from the roots: 2 marks. Cohesion between water molecules keeps the column continuous: 2 marks; adhesion between water molecules and the narrow vessel walls supports the column: 2 marks.
6	Compare the xylem and the phloem in terms of the substances transported, the direction of transport and the structure of the mature cells. (8 marks)	Xylem: transports water and dissolved mineral ions: 1 mark; transport is one way, from root to leaf: 1 mark; mature cells are dead, hollow tubes (vessels or tracheids): 1 mark. Phloem: transports sucrose and amino acids: 2 marks; transport is from source (leaf) to sink (root, fruit, growing point) and can be in either direction: 1 mark; sieve tubes are living cells with sieve plates and companion cells: 2 marks.
7	A plant absorbs 240 cm3 of water in 8 hours. (a) Calculate the mean rate of water uptake in cm3 h-1 and in cm3 min-1. (b) If 228 cm3 is lost by transpiration in the same period, explain what happens to the remaining 12 cm3. (8 marks)	(a) 240 / 8 = 30 cm3 h-1: 2 marks; 30 / 60 = 0.50 cm3 min-1: 2 marks. (b) The remaining 12 cm3: 1 mark; it maintains turgor, supports cell growth and extension, and is a raw material for photosynthesis: 3 marks.
8	Describe the typical leaf symptoms of magnesium deficiency and of nitrogen deficiency, and explain the order in which the symptoms appear in terms of the role of each element. (8 marks)	Magnesium deficiency: older leaves show interveinal yellowing first: 2 marks, because magnesium is the central atom of chlorophyll and is mobile, so the plant moves it to young leaves: 2 marks. Nitrogen deficiency: stunted growth with older leaves turning pale yellow first: 2 marks, because nitrogen is part of proteins, nucleic acids and chlorophyll and is also mobilised from old leaves: 2 marks.

Section C: Experimental Design, Data Analysis and Application (Questions 9-12)

Four questions, 40 marks in total. Suggested time: 25 minutes.

#	Question	Answer and marking notes
9	Design an experiment using an aquatic plant and hydrogencarbonate solution to compare the rate of photosynthesis in red light and in green light. State how variables are controlled and how the rate is measured. (10 marks)	Control: same species and similar length of plant; same concentration of hydrogencarbonate solution; same water temperature maintained by a water bath; same distance from the light source: 4 marks (1 each). Procedure: use red and green filters and count the bubbles released in a fixed time, or collect and measure the gas volume: 3 marks. Repeat each treatment three times and take a mean to reduce random error: 2 marks. Include a dark control to show that the bubbles come from photosynthesis: 1 mark.
10	In an experiment, 24 gas bubbles per minute are released in red light and 6 per minute in green light. (a) Calculate the rate in green light as a percentage of the rate in red light. (b) Explain the difference. (c) Explain why the rate is highest in white light. (10 marks)	(a) 6 / 24 x 100% = 25%: 3 marks. (b) Chlorophyll absorbs mainly red and blue-violet light; most green light is reflected or transmitted, so little is absorbed and the rate is low: 4 marks. (c) White light contains all visible wavelengths, so both the red and the blue-violet bands absorbed by chlorophyll are supplied, giving the greatest total energy absorbed and the highest rate: 3 marks.
11	At noon on a sunny day the carbon dioxide concentration inside a greenhouse falls from 400 ppm to 250 ppm while the rate of photosynthesis drops by about 30 percent. Explain the observation and suggest two practical management measures. (10 marks)	Carbon dioxide has become the limiting factor: 2 marks, because the rate at which photosynthesis removes carbon dioxide exceeds the rate at which ventilation replaces it: 2 marks. Measures: increase ventilation or open the roof vents so outside air enters: 2 marks; add carbon dioxide to about 800 to 1000 ppm: 2 marks; shade or mist the crop to lower the temperature and stop stomata closing: 2 marks.
12	A farmer wants to increase the yield of a leafy vegetable. Using the concept of limiting factors, explain why increasing light intensity alone may not help, and suggest three measures that could raise the rate of photosynthesis. (10 marks)	The rate of photosynthesis is limited by whichever factor is in shortest supply: 3 marks; if light is already sufficient but carbon dioxide or temperature is limiting, extra light gives no further increase: 3 marks. Measures: add carbon dioxide: 1 mark; adjust temperature to the optimum: 1 mark; ensure adequate water and minerals such as magnesium and nitrogen: 1 mark; extend the photoperiod and keep suitable spacing so plants do not shade each other: 1 mark.

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